Nitrogen pentoxide decomposes according to the equation $2\mathrm{N_2O_5(g)}\rightarrow4\mathrm{NO_2(g)}+\mathrm{O_2(g)}$. This first order reaction was allowed to proceed at 40°C and the data below were collected: [N$_2$O$_5$] (M) | Time (min) 0.400 | 0.00 0.289 | 20.0 0.209 | 40.0 0.151 | 60.0 0.109 | 80.0 a) Calculate the rate constant. Include units with your answer. b) What will be the concentration of N$_2$O$_5$ after 100 minutes? c) Calculate the initial rate of the reaction.

CBSE Class 12 · 2011 · Chemical KineticsShort Answer

Nitrogen pentoxide decomposes according to the equation

2 N_{2} O_{5} (g) \to 4 N O_{2} (g) + O_{2} (g)

. This first order reaction was allowed to proceed at 40°C and the data below were collected:

[N

_{2}

_{5}

] (M) | Time (min)
0.400 | 0.00
0.289 | 20.0
0.209 | 40.0
0.151 | 60.0
0.109 | 80.0

a) Calculate the rate constant. Include units with your answer.
b) What will be the concentration of N

_{2}

_{5}

after 100 minutes?
c) Calculate the initial rate of the reaction.

Answer: (a) k ≈ 0.0163 min⁻¹ (b) ≈ 0.079 M (c) ≈ 6.5 × 10⁻³ M min⁻¹

Solution

(a) Using any pair of readings (first order):

k = \frac{2.303}{20} lo g \frac{0.400}{0.289} = \frac{2.303}{20} \times 0.1412

k = 0.0163 mi n^{- 1}

(each interval gives the same value, confirming first order)
(b) At

t = 100

min:

lo g \frac{0.400}{[ R ]} = \frac{0.0163 \times 100}{2.303} = 0.7078 \Rightarrow \frac{0.400}{[ R ]} = 5.10

[N_{2} O_{5}] \approx 0.079 M

= k [N_{2} O_{5}]_{0} = 0.0163 \times 0.400

= 6.5 \times 1 0^{- 3} M mi n^{- 1}

CBSE Class 12 Board · 2011 · Subjective

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CBSE Class 12 · 2011 · Chemical KineticsShort Answer

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